Question: Alain throws a stone off a bridge into a river below. The stone's height (in meters above the water), $x$ seconds after Alain threw it, is modeled by: $h(x)=-5x^2+10x+15$ How many seconds after being thrown will the stone hit the water?
Solution: The stone hits the water when $h(x)=0$. $\begin{aligned} h(x)&=0 \\\\ -5x^2+10x+15&=0 \\\\ x^2-2x-3&=0 \\\\ (x+1)(x-3)&=0 \\\\ \swarrow &\searrow \\\\ x+1=0\text{ or }&x-3=0 \\\\ x=-1\text{ or }&x=3 \end{aligned}$ We found that $h(x)=0$ for $x=-1$ or $x=3$. Since $x=-1$ doesn't make sense in our context, the only reasonable answer is $x=3$. In conclusion, the stone will hit the water after $3$ seconds.